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Inspired by the question

Does the moduli space of smooth curves of genus g contain an elliptic curve

and its amazing answers, I ask (pure out of curiosity) whether the moduli space $M_3$ of (smooth projective connected) curves of genus $3$ contains a (smooth projective connected) curve of genus $2$.

The existence of such a genus two curve is (Edit: stronger) than the existence of a surface $S$, a genus two curve $C$ and a smooth projective non-isotrivial morphism $S\to C$ whose fibres are genus three curves.

If the answer is positive, how explicit can our answer be made? I'm already aware of the fact that $M_g$ contains a complete curve for all $g\geq 3$. For instance, in the paper by Chris Zaal

http://dare.uva.nl/document/38546

many curves of some genus (I think 513) are shown to embed into $M_3$.

Of course, by Shafarevich' conjecture, if $K(C)$ denotes the function field of $C$, there are only finitely many $K(C)$-isomorphism classes of genus three curves over $K(C)$ with good reduction over $C$. I'm asking whether there exists some genus two curve $C$ such that there exists a genus three curve over $K(C)$ with good reduction over $C$.

Edit: the arithmetic analogue also has a negative answer.

The latter (weaker) phrasing of my question allows us to formulate an arithmetic analogue of the above question. (I know that I'm considering function fields over $\mathbf{C}$ and that some of you might argue function fields over $\mathbf{F}_p$ are a better analogue of number fields.) This arithmetic analogue reads as follows. There exists a number field of "genus two" such that there exist a genus three curve over $K$ with good reduction over the ring of integers of $K$. Here a number field of "genus two" should be a number field of absolute discriminant $e^2$. I'll take this to mean discriminant at most $8$.

Arithmetic analogue. (Abrashkin-Fontaine) There do not exist non-zero smooth abelian schemes over the ring of integers of a number field of absolute discriminant at most 8.

There are many related questions I'd also like to ask. For example, what is the minimal $g$ such that $M_g$ contains a genus two curve? Or, what is the minimal $g$ such that $M_3$ contains a genus $g$ curve? And, finally, is there an example of a complete curve in $M_g$ which is defined over $\overline{\mathbf{Q}}$? (Edit: The answer to the last question is positive. This is explained in the comments below.)

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Setting aside the fact that $M_g$ is a stack and not a scheme, non-isotriviality is also too weak: it only says that the map from $C$ to the stack is non-constant, not that it is a closed immersion (or even just injective on geometric points). Also, concerning the question at the very end, by "spreading out and specialization" we see that there are complete curves of a given genus inside $M_g$ over $\overline{\mathbf{Q}}$ if this holds over one algebraically closed field of char. 0 (such as $\mathbf{C}$), or maybe "an example" meant to include "explicit"? –  user27056 Nov 5 '12 at 15:26
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But you're claiming that you're "already aware of the fact that $M_g$ contains a complete curve for all $g \geq 2$". Should that be $>$ instead of $\geq$ then? –  René Nov 5 '12 at 19:42
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@Will Sawin. The locus of hyperelliptic curves (or more generally cyclic covers of prime degree in $M_g$) is affine. This is a result of Gonzalez Diez (1991); see "Loci of curves which are prime Galois coverings of P^1". Probably the result for hyperelliptic curves is more classical. –  Ariyan Javanpeykar Nov 5 '12 at 23:10
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@Will Sawin (continued). So there is no complete curve contained in the locus of hyperelliptic curves of $M_3$. Moreover, the generic fibre of $S\to C$ has to be a smooth quartic. In fact, if it were hyperelliptic, all fibres would be hyperelliptic because $S\to C$ is smooth. –  Ariyan Javanpeykar Nov 5 '12 at 23:13
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@Ariyan. Yes, the result for hyperelliptic curves is very classical. There is an isomorphism $H_g \cong M_{0,2g+2}/\mathbb{S}_{2g+2}$, and $M_{0,n}$ is obviously affine (it is the complement of an arrangement of hyperplanes in $\mathbb{A}^{n-3}$). –  Dan Petersen Nov 6 '12 at 6:29

1 Answer 1

up vote 20 down vote accepted

There does not exist a map of a smooth complete genus 2 curve to $M_3$.

Such a map would give rise to a surface $S$ (of general type) which violates the Bogomolov-Miyaoka-Yau inequality $c_1(S)^2 \leq 3c_2(S)$. This inequality is equivalent to $3\sigma (S) \leq e(S)$ where $\sigma$ and $e$ are the signature and topological euler characteristic of the surface. The euler characteristic of this surface is 8 (since it is multiplicative for fiber bundles) and by the index theorem, the signature is given by 4 times the integral of $\lambda_1$ over the curve in $M_3$. Since $\lambda_1$ is ample on $M_3$, $\sigma$ must be positive and divisible by 4.

This argument is due to Dieter Kotschick in his paper "Signatures, Monopoles, and Mapping Class Groups" (MRL vol 5, 1998).

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Nice.$ $ $ $ $ $ –  J.C. Ottem Nov 6 '12 at 7:34
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A Kodaira fibration (a family of curves corresponding to a nonconstant map $B \to M_g$) always has general type. –  Dan Petersen Nov 6 '12 at 13:45
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Any MO question automatically becomes CW after sufficiently many edits. I do not really understand this feature. –  Dan Petersen Nov 6 '12 at 15:50
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To those interested, I realized that the arithmetic analogue of this question also has a negative answer by a famous theorem of Fontaine-Abrashkin (which I thought applies only to $\mathbf{Z}$ before...) –  Ariyan Javanpeykar Nov 6 '12 at 19:14
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@Olivier : There does indeed exist a genus 2 curve with a non-isotrivial family over it. Moreover, this is the minimal genus with this property. Ron Donagi and I proved that in our Geometry and Topology paper: msp.warwick.ac.uk/gt/2002/06/gt-2002-06-003p.pdf –  Jim Bryan Nov 7 '12 at 0:02

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